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How does one solve an inequality such as the following:

$\sin^{2}(x+y) > \sin^{2}(x) + \sin^{2}(y)$

I can do this:

$(\sin(x)\cos(y) + \sin(y)\cos(x))^{2} > \sin^{2}(x) + \sin^{2}(y)$

$\sin(x)^{2}\cos(y)^{2} + \sin(y)^{2}\cos(x)^{2} + 2\sin(x)\sin(y)\cos(x)\cos(y) > \sin^{2}(x) + \sin^{2}(y)$

$2\sin(x)^{2}\sin(y)^{2} < 2\sin(x)\sin(y)\cos(x)\cos(y)$

The next logical step would be to divide out $\sin(x)\sin(y)$, but I think that causes problems if $\sin(x)\sin(y) < 0$

S L
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RobVerheyen
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    Just tame the "problems". You need to start a case analysis, separately discussing $\sin(x)\sin(y)>0$, $\sin(x)\sin(y)=0$ and $\sin(x)\sin(y)<0$. –  May 13 '14 at 17:35
  • Using http://math.stackexchange.com/questions/175143/prove-sinab-sina-b-sin2a-sin2b or http://math.stackexchange.com/questions/600681/prove-that-sin2a-sin2b-sina-b-sina-b

    $$\sin^2(x+y)-\sin^2x=\sin y\sin(2x+y)$$

    $$\sin^2(x+y)-\sin^2x-\sin^2y=\sin y\sin(2x+y)-\sin^2y=\sin y[\sin(2x+y)-\sin y]$$

    $$=\sin y\cdot\sin x\cdot \cos(x+y)$$

     – lab bhattacharjee May 14 '14 at 18:26
  • Interesting! How do you do the last step though? I seem to be left with more terms after rewriting the second to last expression. – RobVerheyen May 14 '14 at 19:27

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